Question

If one of the lines of the pair $a{x}^2+2hxy+b{y}^2=0$ bisects the angle between positive directions of the axes, then $a,b,\text{and}h$ satisfy the relation

• A. $a+b=2\left|h\right|$
• B. $a+b=-2h$
• C. $a–b=2\left|h\right|$
• D. ${(a–b)}^2=4h^2$

Answer 1

The correct option is B

$a+b=-2h$

Explanation for the correct option

We know that when a pair of a line bisects the angle then the pair can be either $y=x\text{or}y=-x$.

However, it is given that one of the lines bisects the angle in positive directions only. Then the equation will be $y=x$.

Now, apply $y=x$ in the equation $a{x}^2+2hxy+b{y}^2=0$,

$\begin{array}{rcl}\therefore a{y}^2+2h{y}^2+b{y}^2& =& 0\\ & \Rightarrow & y^2\left(a+2h+b\right)=0\\ & \Rightarrow & a+b+2h=0\\ & \Rightarrow & a+b=-2h\end{array}$

Hence, the correct option is (B).